Is 0 Considered a Prime Element in an Integral Domain?

[quasar987]

[SOLVED] Is 0 a prime?

Am I missing something or is 0 a prime element in an integral domain?

In the definition of prime element p of an integral domain, we only ask that the ideal generated by p, be prime.

Well (0) is obviously prime because if ab=0 in an integral domain, then it is that a=0 or b=0.

 

[symbolipoint]

What are the factors of zero?

It might or might not be composite, but it certainly is NOT prime.

 

[quasar987]

I don't understand the point you're making.

 

[CaptainQuasar]

I'm not sure on this but one definition I've heard for a prime is “a number that may only be evenly divided by one and itself.” Zero can be divided by one but dividing by itself would be dividing by zero and hence would be an undefined result.

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[quasar987]

Hi CQ and thanks for your input.

But as far as I can see, this is a definition that makes sense only for integers (and possibly in Euclidean Domains). My point is precisely that the general definition seems to be is in contradiction with this one.

But then again, the general definition claims that -2,-3,-5,... are also primes in Z, while the classical definition considers only positive primes. So it could very well be that 0 is prime in the general setting.

I just wanted to make sure because it surprised me a little to come to this conclusion.

 

[CaptainQuasar]

Hmm… “a number that may only be evenly divided by one and itself and its additive inverse” then, maybe?

Or alternatively it could simply be that zero is an uninteresting prime because it's never a factor of another number.

MathWorld has http://mathworld.wolfram.com/PrimeNumber.html" on prime numbers, BTW.

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[ircdan]

Am I missing something or is 0 a prime element in an integral domain?

In the definition of prime element p of an integral domain, we only ask that the ideal generated by p, be prime.

Well (0) is obviously prime because if ab=0 in an integral domain, then it is that a=0 or b=0.

no definitely not, the definition of a prime element b in a ring R , is that b is a nonzero nonunit element s.t. b |ac => b |a or b| c

 

[quasar987]

Oh yes, I missed the "nonzero" part in the definition. Funny because I re-read it just before posting too.

 

[mathwonk]

all ring elements are divided into three classes: zero divisors (including zero), units (including 1), and the other elements.

included among these other elements are the irreducibles, among which are the primes.

in good cases, all irreducibles are prime, and all these other elements are products of primes.

so primes are never units and never zero divisors, and vice versa.